Question

EXAMPLE 16.5 The large window in Fig. 16-10 is opened using a hydraulic cylinder AB. If the cylinder extends at a constant rate of 0.5 m/s, determine the angular velocity and angular acceleration of the window at the instant 0 30° 1 m SOLUTION Position Coordinate Equation. The angular motion of the window can be obtained using the coordinate 0, whereas the extension or motion along the hydraulic cylinder is defined using a coordinate s, which measures its length from the fixed point A to the moving point B. These coordinates can be related using the law of cosines.mamely 2 m m 2= (2 m)2 + (1 m)2 2(2 m)(1 m) cos 25-4 cos 0 (1) EXAMPLE 16.6 The link shown in Fig. 16-14a is guided by two blocks at A and B, which move in the fixed slots. If the velocity of A is 2 m/s downward, determine the velocity of B at the instant 0 45° 2 m/s 0.2 m SOLUTION I (VECTOR ANALYSIS) Kinematic Diagram. Since points A and B are restricted to move along the fixed slots and v, is directed downward, then velocity va must be directed horizontally to the right, Fig. 16-14b. This motion causes the link to rotate counterclockwise; that is, by the right-hand rule the angular velocity o is dirccted outward, perpendicular to the plane of motion. e45 0.1 m Velocity Equation. Expressing each of the vectors in Fig. 16-14b in terms of their i, j. k components and applying Eq. 16-16 to A, the base EXAMPLE 16.7 The cylinder shown in Fig. 16-15a rolls without slipping on the surface of a conveyor belt which is moving at 2 ft/s. Determine the velocity of point A. The cylinder has a clockwise angular velocity o 15 rad/s at the instant shown. o=15 rad /s 0.5 ft SOLUTION I (VECTOR ANALYSIS) Kinematic Diagram. Since no slipping occurs, point B on the cylinder has the same velocity as the conveyor, Fig. 16-15b. Also, the angular velocity of the cylinder is known, so we can apply the velocity equation to B, the base point, and A to determine vA Velocity Equation VAVB X rA/B (v)i (va)i = 2i (-15k) x (-0.5i +0.5i) = 15 rad/s

Answer 1

16.5

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